Question

A particle carrying 8 electron charges starts from rest and is accelerated through a potential difference of 9000 V. Calculate the KE acquired by it in keV.

Answer 1

q = 8e, u = 0, V = 9000 V, e = 1.6 × 10⁻¹⁹ C

q = 8 (1.6 × 10⁻¹⁹ C)

Initial kE, KE_i = `1.2` mu² = 0

∴ Δ KE = KE_f − KE_i = KE_f

Δ KE = qV

∴ The final KE, KE_f = qV

= (8 × 1.6 × 10⁻¹⁹ C)(9000 V)

= 72 ×1.6 × 10⁻¹⁶ J

= `(72 xx 1.6 xx 10^-16)/(1.6 xx 10^-19)`eV

= 72 × 10³ eV

= 72 keV